Dat Destroyer Organic Chem - Acid Base Extractions lab Question

[Mrhyde]

---

Members do not see this ad.

Number 263.

The explanation for the answer is what I dont understand. My understanding Before reading this explanation:

This is what I had learned before:
for example....
if we need to get R-COOH soluble in water then we use NaHCO3 or NAOH to de-protonate R-COOH to make it R-COO- , so now it can be soluble in water. Then whenever we are done to get back our R-COOH, we Protonate R-COO- , with something like HCl which makes the R-COOH go back so now its not soluble and we can separate now by other means, like filtration.

But this explanation goes against what I just listed above.... What this explanation is saying is
The student ALREADY has hexanoic acid (R-COOH) dissolved in a solution...

1) my first confusion with this is, isen't this wrong since R-COOH cant already be dissolved since its a 6 carbon chain which means that would probably not be soluble without it being de-protonated, despite the smaller COOH polar region since this R group is so large (more then 4 carbons).

2) my next confusion, is then if I assume that its not R-COOH but rather R-COO- that is actually dissolved then the answer should be D) HCl, since we would need to protonate it to get it back to R-COOH which would then form layers for extraction. .....YET the answer is KOH ?

... KOH is like NaOH which would make the R-COOH group into R-COO- ... but ions are what we use to make the substance soluble so this makes no sense to use to separate R-COOH if its already soluble. Since if this question is assuming R-COOH is soluble already (despite the large carbon chain 😵) then KOH would make it more soluble in a sense since it would make it into an ion which dissolves in water lol ? I mean the next question number 264 even confirms this.

I would love for someone to read this explanation in the book (Number 263) and clear it up for 🙂 thank you

 

[Lnguyen7]

Post the question.

 

[Mrhyde]

Post the question.

I dont think i can because it might be a violation of the rules ? so i can only post the book page number

 

[Lnguyen7]

I dont think i can because it might be a violation of the rules ? so i can only post the book page number

Number 263.
Since if this question is assuming R-COOH is soluble already (despite the large carbon chain 😵)

Well then.

What are in the solution? What is the question's question?

From what you posted, it seems the solution contains only organic substances or 1 layer and the task is to extract hexanoic acid. If this is the case, then HA is already soluble because first, other substances are organic and second, the long R chain or like dissolves like. Remember, there is no H2O yet. You are probably confused because of this.

If it was not soluble, then you can just remove the top layer (assuming HA is the least dense) without doing anything further assuming the only thing in that layer is HA.

To extract HA, you would require KOH. KOH will reacts with HA to form a salt and H2O. Now you have two layers: organic and aqueous. The salt dissolves in aqueous layer.